The Stacks project

Separated algebraic spaces are schemes in codimension 1.

Lemma 69.9.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. If $X$ is separated, locally Noetherian, and the dimension of the local ring of $X$ at $x$ is $\leq 1$ (Properties of Spaces, Definition 63.10.2), then there exists an open subspace of $X$ containing $x$ which is a scheme.

Proof. (Please see the remark below for a different approach avoiding the material on finite groupoids.) We can replace $X$ by an quasi-compact neighbourhood of $x$, hence we may assume $X$ is quasi-compact, separated, and Noetherian. There exists a scheme $U$ and a finite surjective morphism $U \to X$, see Limits of Spaces, Proposition 67.16.1. Let $R = U \times _ X U$. Then $j : R \to U \times _ S U$ is an equivalence relation and we obtain a groupoid scheme $(U, R, s, t, c)$ over $S$ with $s, t$ finite and $U$ Noetherian and separated. Let $\{ u_1, \ldots , u_ n\} \subset U$ be the set of points mapping to $x$. Then $\dim (\mathcal{O}_{U, u_ i}) \leq 1$ by Decent Spaces, Lemma 65.12.6.

By More on Groupoids, Lemma 40.14.10 there exists an $R$-invariant affine open $W \subset U$ containing the orbit $\{ u_1, \ldots , u_ n\} $. Since $U \to X$ is finite surjective the continuous map $|U| \to |X|$ is closed surjective, hence submersive by Topology, Lemma 5.6.5. Thus $f(W)$ is open and there is an open subspace $X' \subset X$ with $f : W \to X'$ a surjective finite morphism. Then $X'$ is an affine scheme by Cohomology of Spaces, Lemma 66.17.1 and the proof is finished. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0ADD. Beware of the difference between the letter 'O' and the digit '0'.